What are ' $\mathrm{X}$ ' and ' $\mathrm{Y}$ ' respectively in the following reaction sequence?…
- $\mathrm{X}=\mathrm{PCC}$; $\mathrm{Y}=\text { (i) } \quad \mathrm{B}_2 \mathrm{H}_6$ (ii) $\mathrm{H}_3 \mathrm{O}^{+}$
- $\mathrm{X}=\text { (i) } \mathrm{KMnO}_4 / \mathrm{OH}^{-} ; \quad \mathrm{Y}=\mathrm{NaBH}_4$ (ii) $\mathrm{H}_3 \mathrm{O}^{+}$
- $\mathrm{X}=\text { (i) } \mathrm{KMnO}_4 / \mathrm{OH}^{-}, \Delta ; \quad \mathrm{Y}=\text { (i) } \mathrm{B}_2 \mathrm{H}_6$ (ii) $\mathrm{H}_3 \mathrm{O}^{+}$(ii) $\mathrm{H}_3 \mathrm{O}^{+}$
- $\mathrm{X}=\mathrm{PCC} ; \quad \mathrm{Y}=\mathrm{LAH}$
Solution

$\mathrm{X}=$ (i) $\mathrm{KMnO}_4 / \mathrm{OH}^{-}, \Delta$, (ii) $\mathrm{H}_3 \mathrm{O}^{+}$(for strong side - chain oxidation) $\mathrm{Y}=$ (i) $\mathrm{B}_2 \mathrm{H}_6$, (ii) $\mathrm{H}_3 \mathrm{O}^{+}$(for reduction to alcohols)
Asked in: AP EAMCET 2023 (16 May Shift 2)
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